Generating random graphs in biased Maker-Breaker games

Transcription

1 Generating random graphs in biased Maker-Breaker games Asaf Ferber Michael Krivelevich Humberto Naves August 8, 2014 Abstract We present a general approach connecting biased Maker-Breaker games and problems about local resilience in random graphs. We utilize this approach to prove new results and also to derive some known results about biased Maker-Breaker games. In particular, we show that for b = o ( n), Maker can build a pancyclic graph (that is, a graph that contains cycles of every possible length) while playing a (1 : b) game on E(K n ). As another application, we show that for b = Θ (n/ ln n), playing a (1 : b) game on E(K n ), Maker can build a graph which contains copies of all spanning trees having maximum degree = O(1) with a bare path of linear length (a bare path in a tree T is a path with all interior vertices of degree exactly two in T ). 1 Introduction Let X be a finite set and let F 2 X be a family of subsets. In the (a : b) Maker-Breaker game F, two players, called Maker and Breaker, take turns in claiming previously unclaimed elements of X, with Breaker going first. The set X is called the board of the game and the Institute of Theoretical Computer Science, ETH Zurich, 8092 Zurich, Switzerland. asaf.ferber@inf.ethz.ch. School of Mathematical Sciences, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, 69978, Israel. krivelev@post.tau.ac.il. Research supported in part by USA-Israel BSF Grant and by grant 912/12 from the Israel Science Foundation. Department of Mathematics, UCLA, Los Angeles, CA 90095, USA. hnaves@math.ucla.edu. 1

2 members of F are referred to as the winning sets. Maker claims a board elements per turn, whereas Breaker claims b elements. The parameters a and b are called the bias of Maker and of Breaker, respectively. Maker wins the game as soon as he occupies all elements of some winning set. If Maker does not fully occupy any winning set by the time every board element is claimed by either of the players, then Breaker wins the game. We say that the (a : b) game F is Maker s win if Maker has a strategy that ensures his victory against any strategy of Breaker, otherwise the game is Breaker s win. The most basic case is a = b = 1, the so-called unbiased game, while for all other choices of a and b the game is called a biased game. Note that being the first player is never a disadvantage in a Maker-Breaker game. Therefore, in order to prove that Maker can win some Maker-Breaker game as the first or the second player it is enough to prove that he can win this game as a second player. In this paper we are concerned with providing winning strategies for Maker and hence we will always assume that Maker is the second player to move. It is natural to play Maker-Breaker games on the edge set of a graph G = (V, E). In this case, X = E and the winning sets are all the edge sets of the edge-minimal subgraphs of G which possess some given monotone increasing graph property P. We refer to this game as the (a : b) game P(G). In the connectivity game, Maker wins if and only if his edges contain a spanning tree. In the perfect matching game the winning sets are all sets of V (G) /2 independent edges of G. Note that if V (G) is odd, then such a matching covers all vertices of G but one. In the Hamiltonicity game the winning sets are all edge sets of Hamilton cycles of G. Given a positive integer k, in the k-connectivity game the winning sets are all edge sets of k-vertex-connected spanning subgraphs of G. Given a graph H, in the H-game played on G, the winning sets are all the edge sets of copies of H in G. Playing unbiased Maker-Breaker games on the edge set of K n is frequently in favor of Maker. For example, it is easy to see (and also follows from [21]) that for every n 4, Maker can win the unbiased connectivity game in n 1 moves (which is clearly also the fastest possible strategy). Other unbiased games played on E(K n ) like the perfect matching game, the Hamiltonicity game, the k-vertex-connectivity game and the T -game where T is a spanning tree with bounded maximum degree, are also known to be easy win for Maker (see e.g, [16], [11], [10]). It is thus natural to give Breaker more power by allowing him to claim b > 1 elements in each turn. Note that Maker-Breaker games are known to be bias monotone. That means that none of the players can be harmed by claiming more elements. Therefore, it makes sense to study (1 : b) games and the parameter b which is the critical bias of the game, that is, b is the maximal bias b for which Maker wins the corresponding (1 : b) game F. 2

3 There is a striking relation between the theory of biased Maker-Breaker games and the theory of random graphs, frequently referred to as the Erdős paradigm. Roughly speaking, it suggests that the critical bias for the game played by two clever players and the appropriately defined critical bias for the game played by two random players are asymptotically the same. In this random players version of the game, both players use the random strategy, i.e., Maker claims one random unclaimed element, while Breaker claims b random unclaimed elements from the board E(K n ), per move. Note that the resulting graph occupied by Maker at the end of the game is the random graph G(n, m) with n vertices and m = 1 1+b( n 2) edges. Therefore, if the winning sets consist of all the edge sets of subgraphs of K n which possess some monotone graph property P, a natural guess for the critical bias is b for which m = 1 1+b is the threshold for the property that G(n, m) typically possesses P. For this reason, the Erdős paradigm is also known as the random graph intuition. Chvátal and Erdős were the first to indicate this phenomenon in their seminal paper [9]. They showed that Breaker, playing with bias b = (1+ε)n, can isolate a vertex in Maker s ln n graph while playing on the board E(K n ). It thus follows that Breaker wins every game for which the winning sets consist of subgraphs of K n with positive minimum degree. What is most surprising about their result is that at the end of the game, Maker s graph consists of roughly m = 1 n ln n edges which is (asymptotically) the threshold for a random graph 2 G(n, m) to stop having isolated vertices (for more details on properties thresholds for random graphs, the reader is referred to [7] and [17]). In this spirit, the results of Chvátal and Erdős in [9] hint that b = n is actually the critical bias for many games whose ln n target sets consist of graphs having some property P, for which the threshold probability is p = ln n (such as the connectivity game, the perfect matching game and the Hamiltonicity n game). Gebauer and Szabó showed in [15] that the critical bias for the connectivity game played on E(K n ) is asymptotically equal to n/ ln n. In a relevant development, Krivelevich proved in [18] that the critical bias for the Hamiltonicity game is indeed (1 + o(1))n/ ln n. Another striking result exploring the relation between results in Maker-Breaker games played on graphs and threshold probabilities for properties of random graphs is due to Bednarska and Luczak in [4]. Given a graph G on at least three vertices we define { } E(H) 1 m(g) = max : H G and V (H) 3. V (H) 2 Bednarska and Luczak proved that the threshold bias for the H-game is of order Θ ( n 1/m(H)). The most surprising part in their proof is the side of Maker, where they proved the following: 3

4 Theorem 1.1 (Theorem 2 in [4]). For every graph H which contains a cycle there exists a constant c 0 such that for every sufficiently large integer n and q c 0 n 1/m(H) Maker has a random strategy for the H-game that succeeds with probability 1 o(1) against any strategy of Breaker. Stating it intuitively, they proved that an optimal strategy for Maker is just to claim edges at random without caring about Breaker s moves! Note that since a Maker-Breaker game is a deterministic game, it follows that if Maker has a random strategy that works with non-zero probability against any given strategy of Breaker, then the game is Maker s win (otherwise Maker s strategy should work with probability zero against Breaker s winning strategy). In the proof of Theorem 1.1, the graph obtained by Maker at the end of the game is not exactly a random graph, since some failure edges might exist (that is, it might happen that by choosing random edges, Maker attempts occasionally to pick an edge e which already belongs to Breaker). Thus, in order to prove their result, Bednarska and Luczak not only proved that random graphs typically contain copies of the target graph H, but they also showed that with a positive probability, even after removing a small fraction of the total number of edges, these graphs still contain many copies of H. This particular statement relates to the resilience of random graphs with respect to the property containing a copy of H. Given a monotone increasing graph property P and a graph G which satisfies P, the resilience of G with respect to P measures how much one should change G in order to destroy P (here we assume that an edgeless graph does not satisfy P). There are two natural ways to define it quantitatively. The first one is the following: Definition 1.2. For a monotone increasing graph property P, the global resilience of G with respect to P is the minimum number 0 r 1 such that by deleting r e(g) edges from G one can obtain a graph G not having P. Since one can destroy many natural properties by small changes (for example, by isolating a vertex), it is natural to limit the number of edges touching any vertex that one is allowed to delete. This leads to the following definition of local resilience. Definition 1.3. For a monotone increasing graph property P, the local resilience of G with respect to P is the minimum number 0 r 1 such that by deleting at each vertex v at most r d G (v) edges one can obtain a graph not having P. Sudakov and Vu initiated the systematic study of resilience of random and pseudorandom graphs in [23]. Since then, this field has attracted substantial research interest (see, 4

5 e.g. [2, 5, 6, 8, 14, 19, 20]). Going back to Theorem 1.1, Bednarska and Luczak actually proved that playing according the random strategy, Maker can typically build a graph G G(n, m) minus some ε-fraction of its edges. They then showed that for a given graph H and an appropriate m, the global resilience of a typical G G(n, m) with respect to the property containing a copy of H is at least ε. It is thus natural to seek an alternative theorem which provides the analogous local resilience argument. The main result in this paper uses a sophisticated version of the argument in [4]. Let G be a graph and let 0 < p < 1. The model G(G, p) is a random subgraph G of G, obtained by retaining each edge of G in G independently at random with probability p. For the special case where G = K n, we denote G(n, p) := G(K n, p), which is the well-known Erdős-Rényi model of random graphs. Before stating our main result we need the following definition: Definition 1.4. Let P be a monotone increasing graph property, let G be a graph, and let 0 p, r 1 be two reals. We say that P is (G, p, r)-resilient if the local resilience of a graph G G(G, p) with respect to P is a.a.s. at least r. Our main result is the following. Theorem 1.5. For every constant 0 < ε < 1 and a sufficiently large integer n the following holds. Suppose that (i) 0 p = p(n) 1, and (ii) G is a graph with V (G) = n, and (iii) δ(g) 22 ln n, and εp (iv) P is a monotone increasing graph property which is (G, p, ε)-resilient. ε Then Maker has a winning strategy in the (1 : ) game P(G). 80p As an easy corollary to Theorem 1.5 we establish the following directed analog. A directed graph D consists of a set of vertices V (D), and a set of arcs (or directed edges) E(D) composed of elements of the form (u, v) V (D) V (D), where u v. For a directed graph D and a vertex v V (D) we let d + (v) and d (v) denote the out- and indegrees of v, respectively. Furthermore, we define δ + (D) and δ (D) to be the minimum 5

6 out- and in- degrees of D, respectively, and set δ 0 (D) = min{δ + (D), δ (D)}. Analogously to graphs, we define D(D, p) to be the model of random sub-directed graphs of D obtained by retaining each arc of D with probability p, independently at random. We write D(n, p) for D(D, p) in the special case where D is the complete directed graph on n vertices. That is, V (D) = [n] and E(D) consists of all the possible arcs. For a monotone increasing directed graph property P, we say that P is (D, p, r)-resilient if the local resilience of D with respect to P is at least r, where here we mean that by deleting at each vertex v at most r d + D (v) out- and r d D (v) in-edges one can obtain a directed graph not having P. Theorem 1.6. For every constant 0 < ε < 1 and a sufficiently large integer n the following holds. Suppose that (i) 0 p = p(n) 1, and (ii) D is a directed graph with V (D) = n, and (iii) δ 0 (D) 22 ln n, and εp (iv) P is a monotone increasing graph property which is (D, p, ε)-resilient. ε Then Maker has a winning strategy in the (1 : ) game P(D). 80p Proof. For a directed graph D one can define the following bipartite graph G D : the parts of G D are two disjoint copies of V (D), denoted by A and B. For any a A and b B, the (undirected) edge ab belongs to E(G D ) if and only if the directed edge ab belongs to E(D). Note that the mapping D G D is an injection from the set of all directed graphs on n vertices to the set of bipartite graphs with two parts of size n each, and apply Theorem 1.5 to G D in the obvious way. Note that the property P of digraphs naturally translates to a property P of bipartite graphs which is (K n,n, p, ε)-resilient. Theorems 1.5 and 1.6 connect between Maker s side in biased Maker-Breaker games on graphs or directed graphs and local resilience; it thus allows to use (known) results about local resilience to give a lower estimate for the critical bias in biased Maker-Breaker games. We now present our concrete results for biased games, all of them are applications of Theorems 1.5 and 1.6 and corresponding local resilience results for random graphs. First, as a warm up we prove the following theorem which shows that the critical bias for the Hamiltonicity game played on E(K n ) is Θ( n ln n ). 6

7 Theorem 1.7. There exists a constant δ > 0 for which for every sufficiently large integer n the following holds. Suppose that b δn/ ln n, then Maker has a winning strategy in the (1 : b) Hamiltonicity game played on E(K n ). This result is presented here mainly for illustrative purposes, and also due to the historical importance of the biased Hamiltonicity game and the long road it took before having been resolved finally in [18]. As a second application we show that by playing a (1 : b) game on E(K n ), if b = o( n), then Maker wins the pancyclicity game. That is, Maker can build a graph which consists of cycles of any given length 3 l n. Theorem 1.8. Let b = o( n). Then in the (1 : b) game played on E(K n ), Maker has a winning strategy in the pancyclicity game. Note that the result is asymptotical best possible in a sense that for b 2 n, Chvátal and Erdős showed in [9] that Maker cannot even build a triangle. Ferber, Hefetz, and Krivelevich showed in [12] that if T is a tree on n vertices and (T ) n 0.05 then the following holds. In the (1 : b) game, Maker has a strategy to win the T -game in n + o(n) moves, for every b n They also asked for improvements of the parameter b (regardless of the number of moves required for Maker to win). In this paper, as a third application of our main result, we show how to obtain such an improvement for a large family of trees. Those are trees T with (T ) = O(1) containing a bare path of length Θ(n), where a bare path is a path for which all the interior vertices are of degree exactly two in T. In fact we prove the following much stronger result: Theorem 1.9. For every α > 0 and D > 0 there exists a δ := δ(α, D) > 0 such that for every sufficiently large integer n the following holds. For b δn, in the (1 : b) Makerlog n Breaker game played on E(K n ), Maker has a strategy to build a graph which contains copies of all the spanning trees T such that: (i) (T ) D, and (ii) T admits a bare path of length αn. Remark Note that the bias b in Theorem 1.9 is best possible up to a constant factor, as Chvátal and Erdős showed [9] that for b = (1+ε)n Breaker can isolate a vertex in Maker s ln n graph. 7

8 The rest of the paper is organized as follows: In Section 2 we present some auxiliary results. In Section 3 we prove Theorem 1.5, and in Section 4 we show how to apply Theorem 1.5 combined with local resilience statements (introduced in Subsection 2.3) to various games. 2 Auxiliary results In this section we present some auxiliary results that will be used throughout the paper. 2.1 Binomial distribution bounds We use extensively the following well known bound on the lower and the upper tails of the Binomial distribution due to Chernoff (see, e.g., [1]). Lemma 2.1. If X Bin(n, p), then ( Pr (X < (1 a)np) < exp ( Pr (X > (1 + a)np) < exp a2 np 2 a2 np 3 The following is a trivial yet useful bound. ) for every a > 0. Lemma 2.2. Let X Bin(n, p) and k N.Then Proof. Pr(X k) ( ) n k p k ( ) enp k. k ) for every 0 < a < 1. Pr(X k) ( enp ) k. k 2.2 The MinBox game Consider the following variant of the classical Box Game introduced by Chvátal and Erdős in [9], which we refer to as the MinBox game. The game MinBox(n, D, α, b) is a (1 : b) Maker-Breaker game played on a family of n disjoint sets (boxes), each having size at least D. Maker s goal is to claim at least α F elements from each box F. In the proof 8

9 of our main result, we make use of a specific strategy S for Maker in the MinBox game. This strategy not only ensures his victory, but also allows Maker to maintain a reasonable proportion of elements in all boxes throughout the game. Before describing the strategy, we need to introduce some notation. Assume that a MinBox game is in progress, let w M (F ) and w B (F ) denote the number of Maker s and Breaker s current elements in box F, respectively. Furthermore, let dang(f ) := w B (F ) b w M (F ) be the danger value of F. Finally, we say that a box F is free if it contains an element not yet claimed by either player, and it is active if w M (F ) < α F. Maker s strategy is as follows: Strategy S: In any move of the game, Maker identifies one free active box having maximal danger value (breaking ties arbitrarily), and claims one arbitrary free element from it. We are ready to state the following theorem. Theorem 2.3. Let n, b, and D be positive integers, and 0 < α < 1. Assume that Maker plays the game MinBox(n, D, α, b) according to the strategy S described above. Then he ensures that, throughout the game, every active box F satisfies dang(f ) b(ln n + 1). In particular, if α < 1 game. 1+b b(ln n+1) and D, then S is a winning strategy for Maker in this 1 α(b+1) The proof of this result can be found in the Appendix. We remark that it is very similar to the proof of Theorem 1.2 in [15]. 2.3 Local resilience In this subsection we describe several results related to local resilience of monotone graph properties. The main result of this paper (Theorem 1.5) shows a connection between local resilience of graphs and Maker-Breaker games, therefore, in order to be able to apply it, we first need to present some results related to local resilience of various properties of random graphs. The first statement of this section is a theorem from [20] providing a good bound on the local resilience of a random graph with respect to the property being Hamiltonian. This result will be used in the proof of Theorem 1.7 for the Hamiltonicity game. We remark 9

10 that for our purposes, prior (and weaker) results on the local resilience of a random graph with respect to Hamiltonicity (for example those in [14]) would suffice. Theorem 2.4 (Theorem 1.1, [20]). For every positive ε > 0, there exists a constant C = C(ε) such that for p C ln n, a graph G G(n, p) is a.a.s. such that the following holds. n Suppose that H is a subgraph of G for which G = G \ H has minimum degree at least (1/2 + ε)np, then G is Hamiltonian. The following result from [19] is related to the local resilience of a typical G G(n, p) with respect to pancyclicity. Theorem 2.5 (Thereom 1.1, [19]). If p = ω(n 1/2 ), then the local resilience of G G(n, p) with respect to the property being pancyclic is a.a.s. 1/2 + o(1). The following theorem shows that a sparse random graph G G(n, p) typically contains a copy of every tree T having a bare path of linear length and having bounded maximum degree, even if one deletes a small fixed fraction of edges from each vertex v V (G). This result relates to the local resilience of the property of being universal for this particular class of trees, and it is an essential component in the proof of Theorem 1.9. Theorem 2.6. For every α > 0 and D > 0, there exist ε > 0 and C 0 such that for every p C 0 ln n/n, G G(n, p) is a.a.s. such that the following holds. For every subgraph H G with (H) εnp, the graph G = G \ H contains copies of all spanning trees T such that: (i) (T ) D, and (ii) T contains a bare path of length at least αn. In order to prove Theorem 2.6 we need the following theorem due to Balogh, Csaba and Samotij [2] about the local resilience of random graphs with respect to the property containing all the almost spanning trees with bounded degree. Theorem 2.7 (Theorem 2, [2]). Let β and γ be positive constants, and assume that D 2. There exists a constant C 0 = C 0 (β, γ, D) such that for every p C 0 /n, a graph G G(n, p) is a.a.s. such that the following holds. For every subgraph H of G for which d H (v) (1/2 γ)d G (v) for every v V (G), the graph G = G \ H contains all trees of order at most (1 β)n and maximum degree at most D. 10

11 Proof of Theorem 2.6. Let α > 0 and D > 0 be two positive constants. Let ε := ε(α) > 0 be a sufficiently small constant and let C 0 = C 0 (ε) > 0 be a sufficiently large constant. Let G G(n, p) be a random graph, H G be any subgraph with (H) εnp and denote G = G \ H. We wish to show that G contains a copy of every spanning tree T which satisfies (i) and (ii). This can be done as follows. Assume that G has been generated by a two-round-exposure and is presented as G = G 1 G 2, where G 1, G 2 G(n, q), and 0 q 1 is such that 1 p = (1 q) 2. Note that q > p/2, and since all the properties of G 1 and G 2 that we consider in the proof are monotone increasing, we treat q as p/2. Let V 0 be a random subset of V (G) of size V 0 = 0.99αn and denote G 1 = G 1 [V (G) \ V 0 ]. Note that G 1 G((1 0.99α)n, p/2) and that a.a.s. d G 1 (v) (1 0.99α ε)np/2 for every v V (G 1) (this can be easily shown using Chernoff s inequality, choosing C 0 appropriately, and applying the union bound). Let T be a tree which satisfies (i) and (ii), and let P = v 0 v 1... v t be a bare path of T with t = αn. Let T be the tree obtained from T by deleting v 1,..., v t 1 and adding the edge v 0 v t. Note that V (T ) = (1 α)n + 1. Let β be such that (1 β) V (G 1) = V (T ). Applying Theorem 2.7 to G 1, with (say) γ = 1/4 and β, using the fact that ε is sufficiently small we conclude that there exists a copy T of T in G 1 \ H. Let x and y denote the images (in T ) of v 0 and v t (from T ), respectively. Let V = (V (G) \ V (T )) {x, y}. In order to complete the proof, we should be able to show that (G \ H)[V ] contains a Hamilton path with x and y as its endpoints. Note that V 0 V and that V \ V 0 and the two designated vertices x and y heavily depend on the tree T which we are trying to embed. Therefore, we wish to show that G is a.a.s. such that for every possible option for V (with two designated vertices x and y), (G \ H)[V ] contains a Hamilton path with x and y as its endpoints. For this, first note that since V 0 is a random subset of vertices, V 0 V and G 1 G(n, p/2), using Chernoff s inequality (Lemma 2.1) we conclude that δ ((G 1 \ H) [V ]) 0.491αnp (here we assume that C 0 is large enough). Next, we show that a graph G 1 G(n, p/2) is a.a.s. such that any subgraph D G 1 on αn + 1 vertices with δ(d) 0.49αnp has good expansion properties (our candidate for D will be (G 1 \ H) [V ], and we assume that ε < 0.001α). Claim 2.8. A graph G 1 G(n, p/2) is a.a.s. such that for any subgraph D G 1 with V (D) = αn + 1 and with δ(d) 0.49αnp, the following holds: for every X V (D) with X V (D) /5. N D (X) \ X 2 X + 2 Proof. Let S V (G 1 ) be any subset of vertices of size S 2αn ln n, and note that E G1 (S) 11

12 Bin( ( ) S 2, p/2). Therefore, using Lemma 2.2 we obtain that ( e S 2 p ln ln n Pr( E G1 (S) S np/ ln ln n) 4 S np ( ) S np/ ln ln n e S ln ln n =. 4n ) S np/ ln ln n Applying the union bound we obtain that the probability for having a subset S V (G 1 ) of size S 2αn ln n for which E G1 (S) S np/ ln ln n is at most 2αn ln n s=1 ( n s ) ( ) snp/ ln ln n es ln ln n 4n 2αn ln n s=1 ( en s ) ( ) s snp/ ln ln n es ln ln n = o(1). 4n Now, let D G 1 be a subgraph on αn + 1 vertices with δ(d) 0.49αnp and we wish to show that for every X V (D), if X V (D) /5, then N D (X) \ X 2 X + 2. First, we consider the case X αn 3. Assume that there exists a subset X V (G ln n 1) in this range for which N D (X) \ X 2 X + 1. Using the fact that δ(d) = Θ(np) we obtain E D (X N D (X)) X Θ(np). Now, since E D (X N D (X)) E G1 (X N D (X)) and since X N D (X) αn ln n + 2 < 2αn ln n, we obtain a contradiction. Therefore, we conclude that N D (X) \ X 2 X + 2 holds for every subset X V (D) of size at most αn 3. ln n αn Second, assume 3 < X V (D) /5. In this range it is enough to show that ln n a.a.s. for every two disjoint subsets of vertices X and Y of V (D) of sizes X = αn 3 and ln n Y = αn/10 we have E D (X, Y ) 0. Indeed, let X V (D) be a subset in this range and assume that N D (X)\X 2 X +2. In particular, since X V (D) /5 we conclude that X N D (X) X + 2 X V (D) /5. Therefore, one can easily find X X of size X = αn 3 and Y V (D) \ (X N ln n D(X)) of size Y = αn/10 for which E D (X, Y ) = 0, a contradiction. In order to show that the above mentioned property a.a.s. holds, let X, Y V (G 1 ) be two disjoint subsets of sizes X = αn 3 and Y = αn/10, and observe that E ln n G 1 (X, Y ) Bin( X Y, p/2). Therefore, using Chernoff s inequality we conclude that ( ) Pr E G1 (X, Y ) α2 n 2 p 61 e Θ(n2 p/ ln n). ln n Applying the union bound, we obtain that the probability for having two such subsets 12

13 X and Y with E G1 (X, Y ) α 2 n 2 p/61 ln n is at most ( )( ) n n αn/3 e Θ(n2 p/ ln n) 4 n e Θ(n2 p/ ln n) = o(1). ln n αn/10 Next, let X and Y be two subsets of sizes X = αn 3 ln n and Y = αn/10. Since E G1 (X, Y ) α 2 n 2 p/61 ln n, it follows that the average degree of the vertices of X into Y is at least (3/61) αnp. Therefore, using the facts that ε is much smaller than α, ln ln n that δ(d) 0.49αnp, and that there are at most = o(n/ log n) vertices of degree p larger than (α + ε)np/2 in D (a standard property of random graphs), we conclude that E D (X, Y ) 0. All in all, we conclude that N D (X) \ X 2 X + 2 holds for every X V (D) /5, and it completes the proof. A routine way to turn a non-hamiltonian graph D that satisfies some expansion properties (as in Claim 3.1) into a Hamiltonian graph is by using boosters. A booster is a non-edge e of D such that the addition of e to D creates a path which is longer than a longest path of D, or turns D into a Hamiltonian graph. In order to turn D into a Hamiltonian graph, we start by adding a booster e of D. If the new graph D {e} is not Hamiltonian then one can continue by adding a booster of the new graph. Note that after at most V (D) successive steps the process must terminate and we end up with a Hamiltonian graph. The main point using this method is that it is well known (for example, see [7]) that a non-hamiltonian graph D with good expansion properties has many boosters. However, our goal is a bit different. We wish to turn D into a graph that contains a Hamilton path with x and y as its endpoints. In order to do so, we add one (possibly) fake edge xy to D and try to find a Hamilton cycle that contains the edge xy. Then, the path obtained by deleting this edge from the Hamilton cycle will be the desired path. For that we need to define the notion of e-boosters. Given a graph D and a pair e ( ) V (D) 2, consider a path P of D {e} of maximal length which contains e as an edge. A non-edge e of D is called an e-booster if D {e, e } contains a path P which passes through e and which is longer than P, or that D {e, e } contains a Hamilton cycle that uses e. The following lemma shows that every connected and non-hamiltonian graph D with good expansion properties has many e-boosters for every possible e. Lemma 2.9. Let D be a connected graph for which N D (X)\X 2 X +2 holds for every subset X V (D) of size X k. Then, for every pair e ( ) V (D) 2 such that D {e} does not contain a Hamilton cycle which uses the edge e, the number of e-boosters for D is at least (k + 1) 2 /2. 13

14 The proof of the previous lemma is very similar to the proof of the well known Pósa s lemma using the ordinary boosters ([14], Lemma 4), and hence we postpone it to the appendix. The only difference is that in the proof of Lemma 2.9 we forbid rotations that destroy the edge e; and so the number of possible rotations with a given fixed endpoint drops by at most two. Lastly, we complete the proof of Theorem 2.6 by showing the following lemma, which shows that G 2 contains many boosters. Lemma Assume that G 1 satisfies the condition stated in Claim 2.8. Then G 2 G(n, p/2) is a.a.s. such that for every subgraph H G with (H) εnp the following holds. Suppose that: (i) V V (G) is a subset of size V = αn + 1, and (ii) δ((g 1 \ H)[V ]) 0.49np, and (iii) e = xy ( V 2 ), and (iv) E 0 is a subset of at most αn pairs of V. Then, either G E0,e,V = (G 1\H)[V ] E 0 {e} contains a Hamilton cycle that passes through e, or G 2 contains at least α 2 n 2 /100 e-boosters for G E0,e,V. Proof. Using Lemma 2.9 we conclude that for every such V, e ( ) V 2, and for every subset E 0 of at most αn pairs of V, G E0,e,V has at least α2 n 2 /3 e-boosters. For a fixed choice of such V, e and E 0, by Chernoff s inequality (Lemma 2.1) it follows that the probability that G 2 will have at most α 2 n 2 p/100 e-boosters for G E0,e,V is at most exp( Cn2 p), where C is a constant which depends only in α. Applying the union bound, running over all the options for choosing H, e and E 0 we obtain that the probability that there exist such V, e, and E 0 for which G 2 contains at most α 2 n 2 p/100 e-boosters is at most εn 2 p ( ( e(g1 ) α )n 2 2 n 2 t αn t=1 ( ( εn 2 e(g1 ) α p )n 2 2 n 2 εn 2 p αn ( eα 2 n 2 p εn 2 p εn 4 p ) exp( Cn 2 p) ) exp( Cn 2 p) ) εn 2 p (eαn) αn exp( Cn 2 p) 14

15 ( ) eα εn 4 2 εn 2 p p (eαn) αn exp( C C 0 n ln n) = o(1), ε where the last inequality holds since ε is much smaller than α and since C 0 is large enough. This completes the proof of Theorem Proof of the main result Proof of Theorem 1.5. In order to prove the theorem, we provide Maker with a random strategy that enables him to generate a random graph G G(G, p), and claim at least 1 ε fraction of the edges of G touching each vertex. We then use the fact that P is (G, p, ε) resilient to conclude that G a.a.s. satisfies P. Note that since a Maker-Breaker game is deterministic, and since the strategy we describe a.a.s. ensures Maker s win against any strategy of Breaker, it follows that Maker also has a deterministic winning strategy. We now present the random strategy for Maker. In this strategy, Maker will gradually generate a random graph G G(G, p), by tossing a biased coin on each edge of G, and declaring that it belongs to G independently with probability p. Each edge which Maker has tossed a coin for is called exposed, and we say that Maker is exposing an edge e E(G) whenever he tosses a coin to decide about the appearance of e in G. To keep track of the unexposed edges, Maker maintains a set U v N G (v) of the unexposed neighbors of v, for each vertex v in G; i.e. u U v if and only if the edge vu remains to be exposed. Initially, U v = N G (v) for all v V (G). We remark that Maker will expose all edges of G, even those that belong to Breaker. In every turn, Maker chooses an exposure vertex v (we will later discuss the choice of the exposure vertex) and starts to expose edges connecting v to vertices in U v, one by one in an arbitrary order, until one edge in G is found (that is, until he has a first success). If this exposure happens to reveal an edge vu E(G ) not yet claimed by Breaker, Maker claims it and completes his move. Otherwise, either the exposure failed to reveal a new edge in G (failure of type I ), or the newly found edge already belongs to Breaker (failure of type II ). In either case, Maker skips his move. Let f I (v) and f II (v) denote the number of failures of type I and II, respectively, for the exposure vertex v. To complete the proof, it suffices to show that a.a.s. Maker can ensure that f II (v) εd G (v)p for all vertices v V (G) at the end of the game (note that if a failure of type I occurs, it does not harm Maker in claiming edges of the generated random graph G ). 15

16 To keep the failures of type II under control, concurrently to the game played on G, we simulate a game MinBox(n, 4δ(G), p/2, 2b). In this simulated game, there is one box F v for each v V (G) which helps us to keep track of the exposure of edges touching v. Initially, we set the sizes of the boxes as F v = 4d G (v). Now, we describe Maker s strategy. Maker s strategy S M : Maker s strategy is divided into the following two stages. Stage 1: Before his move, Maker updates the status of the simulated game by pretending that Breaker claimed one free element from both F v and F u, for each edge vu occupied in Breaker s last move. Maker then identifies a free active box F v having highest danger value in the simulated game (breaking ties arbitrarily). If there is no such box, Maker proceeds to the second stage of the strategy. Otherwise, let F v be such a box. Maker claims one free element from F v, and selects v as the exposure vertex. Let σ : [m] U v be an arbitrary permutation on U v, where m := U v. Maker starts tossing a biased coin for vertices in U v, independently at random, according to the ordering of σ. (a) If there were no successes, then Maker declares this turn as a failure of type I, thereby incrementing f I (v), and skips his move in the original game. Maker then claims p 2 F v 1 additional free elements from F v in the simulated game, and updates U v :=, and U σ(i) := U σ(i) \ {v} for each i m. Assume that Maker s first success has happened at the kth coin tossing. (b) If the edge vσ(k) is not free, then Maker declares vσ(k) as a failure of type II, increments f II (v) by one, and skips his move in the original game. Maker then updates U v := U v \ {σ(i) : i k}, and U σ(i) := U σ(i) \ {v} for each i k. (c) Otherwise, Maker claims the edge vσ(k). In this case Maker also claims a free element from box F σ(k) and then updates U v := U v \ {σ(i) : i k}, and U σ(i) := U σ(i) \ {v} for each i k. Stage 2: In this stage, there are no free active boxes. Let U := {vu : v V (G), u U v }. For each e = vu U, Maker declares a failure of type II on both u and v (i.e., increments both f II (u) and f II (v) by one) with probability p, independently at random. After the end of this stage, Maker stops playing the game altogether, and skips all his subsequent moves. We now prove that by following S M, Maker typically achieves his goal. For the sake of notation, at any point during the game, we denote by d M (v) and d B (v) the degrees of v in the subgraphs currently occupied by Maker and Breaker, respectively. The proof will follow from the next four claims. 16

17 Claim 3.1. At any point during the first stage, we have w M (F v ) (1 + 2p)d G (v) and w B (F v ) d G (v) for every box F v in the simulated game. In particular, no box is ever exhausted of free elements. Proof. Clearly w B (F v ) = d B (v) d G (v). Moreover, w M (F v ) = d M (v) + p 2 F v f I (v) + f II (v). Since d M (v) + f II (v) d G (v) and f I (v) 1, as F v becomes inactive after a failure of type I on v, we have w M (F v ) d G (v) + p 2 F v (1 + 2p)d G (v), as required. Claim 3.2. For every v V (G), F v becomes inactive before d B (v) εd G (v)/5. Proof. Let v V (G) be any vertex of V (G). Since in the simulated game Maker follows the strategy described in Theorem 2.3, we must have dang(f v ) = w B (F v ) 2b w M (F v ) 2b(ln n + 1) (1) for every active box F v. Assume that there exists a vertex v V (G) for which F v is still active and w B (F v ) = d B (v) εd G (v)/5. Recall that b = ε, and by (1) it follows that 40p 2 ε 40p (ln n + 1) w B(F v ) 2 ε 40p w M(F v ) εd G (v)/5 ε 20p w M(F v ). Therefore, we obtain that w M (F v ) 4d G (v)p (ln n + 1), and by the assumption that 11 ln n δ(g), we conclude that w εp M (F v ) > 3.8d G (v)p p F 2 v. However, since we assumed that F v is active in MinBox(n, 4δ(G), p/2, 2b), we must have w M (F v ) p F 2 v, which is a contradiction. Claim 3.3. All edges of G are a.a.s. exposed before the beginning of Stage 2. Proof. Suppose there exists a vertex v at the beginning of the second stage, such that U v. Since U v, we must have f I (v) = 0. Moreover, because F v is not active, we must also have w M (F v ) = d M (v) + f II (v) p F 2 v = 2d G (v)p. This implies that d G (v) d M (v) + f II (v) 2d G (v)p. Now, since d G (v) Bin(d G (v), p), using Chernoff s inequality, it follows that ( ) 1 P(Bin(d G (v), p) 2d G (v)p) < e dg(v)p/4 = o. n Applying the union bound, it thus follows that with probability 1 o(1), there exists no such vertex, proving the claim. 17

18 Claim 3.4. For every v V (G) a.a.s. f II (v) εd G (v)p. Proof. Let v V (G) be any vertex. By Claim 3.2, during Stage 1 Breaker can touch v at most εd G (v)/5 times before F v becomes inactive. Since a failure of type II occurs if Maker has a success on one of Breaker s edges, it follows that f II (v) is stochastically dominated by Bin(m, p), where m εd G (v)/5. Applying Lemma 2.2 to f II (v) we conclude that the probability for having more than εd G (v)p edges vu which are failures of type II is at most P (Bin(εd G (v)/5, p) εd G (v)p) ( ) εdg (v)p ( ) eεdg (v)p/5 1 = o. εd G (v)p n Applying the union bound we obtain that the probability that there is such a vertex is o(1). Moreover, by Claim 3.3, with probability 1 o(1) all the edges of G were exposed before the beginning of Stage 2. Therefore, a.a.s. f II (v) εd G (v)p for all v V (G). This completes the proof of Theorem Applications In this section we show how to apply Theorems 1.5 and 1.6 in order to prove Theorems 1.7, 1.8 and 1.9. We start with proving Theorem 1.7, which states that Maker can win the Hamiltonicity game played on E(K n ) against an asymptotically optimal (up to a constant factor) bias of Breaker. Proof of Theorem 1.7. Let C 1 = C( 1) be as in Theorem 2.4, and let C 6 2 := max{c 1, 1000}. First, observe that for p C 2 ln n a.a.s. we have that G G(n, p) satisfies δ(g) 5 np (this n 6 follows immediately from Chernoff and the union bound). Next, note that the property P := being Hamiltonian is (K n, p, 1/6) resilient for p C 2 ln n. Indeed, let H G be a n subgraph for which d H (v) 1d 6 G(v). Observe that in G := G H we have d G (v) 5d 6 G(v). Now, since a.a.s. δ(g) 5np, it follows that 6 δ(g ) 25np > 2 np. Therefore, by our choice 36 3 of C 2 and Theorem 2.4, it follows that G is Hamiltonian. Lastly, applying Theorem 1.5 with ε = 1, K 6 n (as the host graph G), p = C 2 ln n and n 1 P, we obtain that Maker has a winning strategy in the (1 : ) game P(K 480p n). Note that 1 = n 1 480p 480C 2, and therefore, by setting δ := ln n 480C 2 we complete the proof. Next, we prove Theorem

19 Proof of Theorem 1.8. Let p = ω(n 1/2 ) and note that by Theorem 2.5, it follows that the property P := being pancyclic is (K n, p, 1/2 + o(1))-resilient. Therefore, by applying Theorem 1.5 with (say) ε = 1/3, K n (As the host graph), p and P, we obtain that Maker has a winning strategy in the (1 : 1 480p ) game P(K n). This completes the proof. Finally, we prove Theorem 1.9. Proof of Theorem 1.9. Let α > 0 and D > 0 be two positive constants. Let ε > 0 and C 0 be as in Theorem 2.6 (applied to α and D). Let C 1 max { } C 0, 44 ε be a large enough constant for which G G(n, p) a.a.s. satisfies (G) (1 + ε)np, provided that p C 1 ln n. n Let T be the set of all trees T on n vertices satisfying: (i) (T ) D, and (ii) T contains a bare path of length at least αn, and let P be the property being T -universal (that is, contains copies of all trees in T ). Observe that P is ( ( ) ε K n, p, 1+ε) resilient, and hence Kn, p, ε 2 resilient for every p C 1 ln n. n Indeed, let H be a subgraph of G for which d H (v) ε d 1+ε G(v), for all vertices v V (G). Thus d H (v) ε (G) εnp. Therefore, by Theorem 2.6, 1+ε G := G \ H satisfies P. Lastly, by applying Theorem 1.5 with ε/2 (as ε), K n (as the host graph G), p, and P, ε we obtain that Maker has a winning strategy in the (1 : ) game P(K 160p n). By setting δ = ε 160C 1, we complete the proof. Note that we used the fact that C 1 44 in order to ε verify assumption (iii) in Theorem 1.5. Acknowledgment. The authors wish to thank the anonymous referees for many valuable comments. References [1] N. Alon and J. H. Spencer, The Probabilistic Method, Wiley, New-York, [2] J. Balogh, B. Csaba and W. Samotij, Local resilience of almost spanning trees in random graphs, Random Structures and Algorithms 38, no. 1-2 (2011), [3] J. Beck, Combinatorial Games: Tic-Tac-Toe Theory, Cambridge University Press,

20 [4] M. Bednarska and T. Luczak, Biased positional games for which random strategies are nearly optimal, Combinatorica 20 (2000), [5] S. Ben-Shimon, M. Krivelevich and B. Sudakov, Local resilience and Hamiltonicity Maker-Breaker games in random regular graphs, Combinatorics, Probability and Computing 20 (2011), [6] S. Ben-Shimon, M. Krivelevich and B. Sudakov, On the resilience of Hamiltonicity and optimal packing of Hamilton cycles in random graphs, SIAM Journal on Discrete Mathematics 25 (2011), [7] B. Bollobás, Random Graphs, Cambridge University Press, [8] J. Böttcher, Y. Kohayakawa, and A. Taraz, Almost spanning subgraphs of random graphs after adversarial edge removal, Electronic Notes in Discrete Mathematics 35 (2009), [9] V. Chvátal and P. Erdős, Biased positional games, Annals of Discrete Mathematics 2 (1978), [10] D. Clemens, A. Ferber, R. Glebov, D. Hefetz and A. Liebenau, Building spanning trees quickly in Maker-Breaker games, arxiv preprint arxiv: (2013). [11] A. Ferber and D. Hefetz, Weak and strong k-connectivity game, European Journal of Combinatorics (2014), pp [12] A. Ferber, D. Hefetz and M. Krivelevich, Fast embedding of spanning trees in biased Maker-Breaker games, European Journal of Combinatorics 33 (2012), [13] A. Frieze, An algorithm for finding hamilton cycles in random digraphs, Journal of Algorithms 9 (1988), [14] A. Frieze and M. Krivelevich, On two Hamilton cycle problems in random graphs, Israel Journal of Mathematics 166 (2008), [15] H. Gebauer and T. Szabó, Asymptotic random graph intuition for the biased Connectivity game, Random Structures and Algorithms, 35 (2009), [16] D. Hefetz, M. Krivelevich, M. Stojakovic and T. Szabó, Fast winning strategies in Maker-Breaker games, Journal of Combinatorial Theory Series B 99 (2009), [17] S. Janson, T. Luczak and A. Ruciński, Random graphs, Wiley, New York,

21 [18] M. Krivelevich, The critical bias for the Hamiltonicity game is (1+o(1))n/ ln n, Journal of the American Mathematical Society 24 (2011), [19] M. Krivelevich, C. Lee and B. Sudakov, Resilient pancyclicity of random and pseudorandom graphs, SIAM Journal on Discrete Mathematics 24 (2010), [20] C. Lee and B. Sudakov, Dirac s theorem for random graphs, Random Structures and Algorithms 41, no. 3 (2012), [21] A. Lehman, A solution to the Shannon switching game, J. Soc. Indust. Appl. Math. 12 (1964), [22] L. Pósa, Hamiltonian circuits in random graphs, Discrete Math. (1976) 14, [23] B. Sudakov and V. H. Vu, Local resilience of graphs, Random Structures and Algorithms 33, no. 4 (2008), [24] D. B. West, Introduction to Graph Theory, Prentice Hall, A Proofs of Theorem 2.3 and Lemma 2.9 We begin with the proof of the MinBox game. Proof of Theorem 2.3. The proof of this theorem is very similar to the proof of Theorem 1.2 in [15]. Since claiming an extra element is never a disadvantage for any of the players, we can assume that Breaker is the first player to move. For a subset X of boxes, let F X dang(x) = dang(f ) denote the average danger of the boxes in X. The game ends when X there are no more free elements left. First we prove the upper bound for the danger values of active boxes. Suppose, towards a contradiction, that there exists a strategy for Breaker that ensures the existence of an active box F satisfying dang(f ) > b(ln n + 1) at some point during the game. Denote the first time when this happens by g. Let I = {F 1,..., F g } be the set which defines Maker s game, i.e, in his i th move, Maker plays at F i for 1 i g 1 and F g is the first active box satisfying dang(f g ) > b(ln n + 1). For every 0 i g 1, let I i = {F g i,..., F g }. Following the notation of [15], let dang Bi (F ) and dang Mi (F ) denote the danger value of a box F, directly before Breaker s and Maker s i th move, respectively. Notice that in his g th move, Breaker increases the danger value of F g to more than b(ln n + 1). This is only possible if dang Bg (F g ) > b(ln n + 1) b = b ln n. 21

22 Analogously to the proof of Theorem 1.2 in [15], we state the following lemmas which estimate the change of the average danger after a particular move (by either player). In the first lemma we estimate the changes after Maker s moves. Lemma A.1. Let i, 1 i g 1, (i) if I i I i 1, then dang Mg i (I i ) dang Bg i+1 (I i 1 ) 0. (ii) if I i = I i 1, then dang Mg i (I i ) dang Bg i+1 (I i 1 ) b I i. Proof. For part (i) we have that F g i I i 1. Since danger values do not increase during Maker s move, we have dang Mg i (I i 1 ) dang Bg i+1 (I i 1 ). Before M g i, Maker selected the box F g i because its danger was highest among the active boxes. Thus dang(f g i ) max(dang(f g i+1 ),..., dang(f g )), which implies dang Mg i (I i ) dang Mg i (I i 1 ). Combining the two inequalities establishes part (i). For part (ii) we have that F g i I i 1. In M g i, w M (F g i ) increases by 1 and w M (F ) does not change for any other box F I i. Besides, the values of w B ( ) do not change during Maker s move. So dang(f g i ) decreases by b, whereas dang(f ) do not increase for any other box F I i. Hence dang(i i ) decreases by at least In the second lemma we estimate the changes after Breaker s moves. Lemma A.2. Let i be an integer, 1 i g 1. Then, dang Mg i (I i ) dang Bg i (I i ) b I i b I i., which implies (ii). Proof. The increase of F I i w B (F ) during B g i is at most b. Moreover, since the values of w M (F ) for F I i do not change during Breaker s move, the increase of dang(i i ) (during B g i ) is at most, which establishes the lemma. b I i Combining Lemmas A.1 and A.2, we obtain the following corollary which estimates the change of the average danger after a full round. Corollary A.3. Let i be an integer, 1 i g 1. (i) if I i = I i 1, then dang Bg i (I i ) dang Bg i+1 (I i 1 ) 0. (ii) if I i I i 1, then dang Bg i (I i ) dang Bg i+1 (I i 1 ) b I i 22

23 Next, we prove that before Breaker s first move, dang B1 (I g 1 ) > 0, thus obtaining a contradiction. To that end, let I g 1 = r and let i 1 <... < i r 1 be those indices for which I ij I ij 1. Note that I ij = j + 1. Recall that dang Bg (F g ) > b ln n, therefore g 1 ( ) dang B1 (I g 1 ) = dang Bg (I 0 ) + dang Bg i (I i ) dang Bg i+1 (I i 1 ) i=1 r 1 dang Bg (I 0 ) + dang Bg (I 0 ) j=1 r 1 j=1 ( ) dang Bg ij (I ij ) dang (I Bg ij +1 i j 1) b j + 1 dang Bg (I 0 ) b ln n > 0, [by Corollary A.3(ii)] [by Corollary A.3(i)] and this contradiction establishes the upper bound for the danger values of active boxes. Lastly, consider a MinBox(n, D, α, b) game where α < 1 b(ln n+1) and D. We will b+1 1 α(b+1) prove that S is a winning strategy for Maker in this setting. With this in mind, it suffices to show that there are no active boxes left at the very end of the game. Suppose not, and let F be a box which remained active, i.e., w M (F ) < α F. Clearly F is not free, since the game has ended. Thus we have w M (F ) + w B (F ) = F. Moreover, since Maker played according to S, we must have dang(f ) b(ln n + 1). Hence b(ln n + 1) dang(f ) = w B (F ) b w M (F ) = F (b + 1)w M (F ) > (1 α(b + 1)) F. b(ln n+1) This implies that D F <, which is a contradiction, thereby proving that Maker 1 α(b+1) is the winner, and concluding the proof of the theorem. We turn to prove the variant of Pósa s lemma for e-boosters. Proof of Lemma 2.9. Let D be a connected graph for which N D (X) \ X 2 X + 2 holds for every subset X V (D) of size X k. Let e ( ) V (D) 2 be a pair such that the graph D {e} does not contain a Hamilton cycle which uses e. We will prove that the number of e-boosters for D is at least (k + 1) 2 /2. The idea behind the proof is fairly natural and is based on Pósa s rotation-extension technique. Let P = x 0 x 1... x h be a path in D {e}, starting at a vertex x 0. Suppose P contains e, say e = x i x i+1 for some 0 i < h. If D contains an edge x j x h for some 23

24 0 j < h 1 such that j i, then one can obtain a new path P of the same length as P which contains e. The new path is P = x 0 x 1... x j x h x h 1... x j+1, obtained by adding the edge x j x h and deleting x j x j+1. This operation is called an elementary rotation at x j with fixed x 0. We can therefore apply other elementary rotations repeatedly, and if after a number of rotations, an endpoint x of the obtained path Q is connected by an edge to a vertex y outside Q, then Q can be extended by adding the edge xy. The power of the rotation-extension technique of Pósa hinges on the following fact. Let P = x 0... x h be a longest path in D {e} containing e. Let P be the set of all paths obtainable from P by a sequence of elementary rotations with fixed x 0. Denote by R the set of the other endpoints (not x 0 ) of paths in P, and by R and R + the sets of vertices immediately preceding and following the vertices of R along P, respectively. We claim that: Claim A.4. N D (R) \ R R R + e. Proof of Claim A.4. Fix u R, let v V (D) \ (R R R + e), and consider a path Q P ending at u. If v V (D) \ V (P ), then uv E(D), as otherwise the path Q can be extended by adding v, thus contradicting our assumption that P is a longest path in D {e} containing e. Suppose now that v V (P ) \ (R R R + e). Then v has the same two neighbors in every path in P, because an elementary rotation that removed one of its neighbors along P would, at the same time, put either this neighbor or v itself in R (in the former case v R R + ). Then if u and v are adjacent, an elementary rotation at v can be applied to Q (since v e), and produces a path in P whose endpoint is a neighbors of v along P, a contradiction. Therefore in both cases u and v are non-adjacent, thereby proving Claim A.4. Equipped with Claim A.4 we turn back to the proof of the lemma. Again, let P = x 0 x 1... x h be a longest path in D {e} containing e, and let R, R, R + be as in Claim A.4. Note that R R and R + R 1, since x h R has no following vertex on P, and thus does not contribute an element to R +. According to Claim A.4, we have N D (R) \ R R R + e 2 R + 1, and it follows that R > k. We claim that, for each v R, the pair x 0 v is an e-booster for D. To prove this claim, fix v R, and let Q P be a path ending at v. Note that by adding x 0 v to Q, we turn Q into a cycle C containing e. This cycle is either Hamiltonian or V (Q) V (D). The former case would immediately imply that x 0 v is an e-booster for D. Thus we may assume that V (C) = V (Q) V (D). Since D is connected, there exists 24